The correct statement for the molecule CsI$$_3$$, is:

We begin with the observation that caesium, $$Cs$$, belongs to the alkali-metal group. An alkali metal almost always loses a single electron to attain the noble-gas configuration, so the most stable and common oxidation state for caesium is $$+1$$. Thus the ionic form of caesium that we expect is $$Cs^+$$.

Now let us look at the iodine part. If the empirical formula of the solid is written as $$CsI_3$$, we must decide how the three iodine atoms combine with the single caesium ion. Let us denote the overall charge on the iodine-containing species as $$x$$. Because a neutral compound has total charge $$0$$, we can write a simple charge-balance equation:

$$\text{(charge on }Cs) + \text{(charge on iodine species)} = 0$$

Substituting the known charge $$+1$$ for $$Cs^+$$ gives

$$+1 + x = 0.$$

Solving for $$x$$, we move $$+1$$ to the right:

$$x = -1.$$

Hence the iodine-containing unit must carry a net charge $$-1$$. The simplest and most familiar polyiodide anion of charge $$-1$$ is the triiodide ion $$I_3^-$$. In fact this ion is well known: when a metal iodide such as $$KI$$ or $$CsI$$ is treated with molecular iodine $$I_2$$, the following equilibrium is established:

$$I^- + I_2 \rightleftharpoons I_3^-.$$

Replacing $$I^-$$ by $$CsI$$, we can depict the actual formation of the solid as

$$CsI + I_2 \rightarrow CsI_3 \;(\text{containing } Cs^+ \text{ and } I_3^-).$$

This clearly shows that the crystal of $$CsI_3$$ is best described as an ionic compound made up of $$Cs^+$$ cations and $$I_3^-$$ anions.

Let us quickly examine the alternative statements to ensure they are incorrect:

Option A: “It is a covalent molecule.” — The presence of a classical alkali-metal cation and a polyiodide anion means the bonding is predominantly ionic, not purely covalent.

Option C: “It contains $$Cs^{3+}$$ and $$I^-$$ ions.” — Caesium does not normally exhibit the $$+3$$ oxidation state, and the charge balance $$(+3) + (-1\!\times\!3) = 0$$ would actually fit $$CsI_3$$ stoichiometry, but such a high oxidation state is chemically unrealistic for $$Cs$$.

Option D: “It contains $$Cs^+$$, $$I^-$$ and lattice $$I_2$$ molecule.” — If the solid contained separate $$I^-$$ and $$I_2$$ species, the stoichiometric ratio would still be $$1:1:1$$, yet diffraction and spectroscopic data reveal the linear polyiodide ion $$I_3^-$$ rather than discrete $$I^-$$ plus molecular $$I_2$$.

Only Option B correctly reflects the ionic composition and charge balance.

Hence, the correct answer is Option B.